Solve for $x$ and $y$ using substitution. ${-4x+5y = -7}$ ${x = 4y+10}$
Since $x$ has already been solved for, substitute $4y+10$ for $x$ in the first equation. ${-4}{(4y+10)}{+ 5y = -7}$ Simplify and solve for $y$ $-16y-40 + 5y = -7$ $-11y-40 = -7$ $-11y-40{+40} = -7{+40}$ $-11y = 33$ $\dfrac{-11y}{{-11}} = \dfrac{33}{{-11}}$ ${y = -3}$ Now that you know ${y = -3}$ , plug it back into $\thinspace {x = 4y+10}\thinspace$ to find $x$ ${x = 4}{(-3)}{ + 10}$ $x = -12 + 10$ ${x = -2}$ You can also plug ${y = -3}$ into $\thinspace {-4x+5y = -7}\thinspace$ and get the same answer for $x$ : ${-4x + 5}{(-3)}{= -7}$ ${x = -2}$